TITLE: Vector Calculus Identities Using Differential Forms
QUESTION [17 upvotes]: Is there a nice way to derive
$$ \nabla (\vec{F} \cdot \vec{G}) = \vec{F} \times (\nabla \times \vec{G} ) + \vec{G} \times (\nabla \times \vec{F}) + (\vec{F} \cdot \nabla ) \vec{G} + (\vec{G} \cdot \nabla ) \vec{F} $$
&
$$ \nabla \times (\vec{F} \times \vec{G}) = ( \vec{G} \cdot \nabla )\vec{F} - (\vec{F} \cdot \nabla ) \vec{G} + \vec{F}( \nabla \cdot \vec{G} ) - \vec{G} ( \nabla \cdot \vec{F}) $$
using differential forms?

REPLY [8 votes]: I think I have a way to do it using differential forms operations.  
First a quick note on notation, I'll use $F$ to denote both the differential one form $F_a$ as well as the vector $F^a = g^{ab} F_b$.  Hopefully there will be no confusion; mostly I'll use  $F$ and $G$ as vectors when doing Lie derivatives, $£_F$, or taking inner products $i_F$.  
The tricky part was to come up with a way to deal with $(F\cdot\nabla)G$ and similar terms.  For the first identity, we can begin by noticing that $(F\cdot\nabla)G = F^a\nabla_a G_b$ looks like the first term in the expression for the Lie derivative of a 1-form:
$$£_F G = F^a\nabla_a G_b + G_a\nabla_bF^a.$$
The other term looks like one of the terms you get when you take the gradient of $F^a G_a$, i.e.
$$\nabla_b(F^a G_a)=G_a\nabla_b F^a + F_a\nabla_b G^a.$$
So it is clear if we switch the roles of $F$ and $G$, we will get the other part of this term, so we arrive at the necessary identity,
$$£_F G + £_G F -d(i_F G) = (F\cdot\nabla)G+(G\cdot\nabla)F. $$
Also note the following identities which are fairly straightforward to prove:
$$dG = *(\nabla\times G)$$
$$i_F*H = *(H\wedge F)=-F\times H,$$
from which it follows
$$-i_FdG = *(F\wedge * dG)= F\times(\nabla\times G) $$
Using these, we can prove the first identity, invoking Cartan's magic formula, $£_F  = i_Fd+di_F $:
$$d(i_F G) = -i_FdG+£_FG $$
$$d(i_G F) = -i_GdF+£_GF $$
Since $i_F G = i_G F = F\cdot G$, we can add these expressions to get
$$ 2d(i_F G) = -i_FdG-i_GdF + £_FG+£_GF$$
$$\boxed{d(i_FG) = *(F\wedge*dG)+*(G\wedge*dF)+£_FG+£_GF-d(i_FG)} $$
This is the differential forms version of your first identity (the LHS is the gradient of the scalar $i_FG = F\cdot G$).  Note that the appearance of $d(i_F G)$ on the RHS means that this expression has a more natural representation by differential forms by moving that term back over to the left, i.e.
$$ d(i_F G) = \frac12(*(F\wedge*dG)+*(G\wedge*dF) + £_FG+£_GF)$$
The second identity is actually a bit cleaner.  We begin by noting that for a vector $F^a$, we can represent the dual of the associated one form $F_b$ by
$$*F = i_F \epsilon,$$
where $\epsilon$ is the volume $n$-form ($n=3$ for us).  Using our previous identities, we can determine that the LHS is $*d*(F\wedge G)$.  So we have
\begin{align}
*d*(F\wedge G)& = *di_G*F \\
&= *(£_G*F-i_Gd*F)\\
&=*(£_Gi_F\epsilon - i_Gd*F)\\
&=*(i_{[G,F]}\epsilon+i_F£_G\epsilon - i_Gd*F)\\
&=*(i_{[G,F]}\epsilon+i_Fdi_G\epsilon+i_Fi_Gd\epsilon-i_Gd*F)
\end{align}
$$\boxed{*d*(F\wedge G) = [G,F]^\flat +(*d*G)\wedge F - (*d*F)\wedge G }$$
I put a $\flat$ on the commutator of the vectors $[G,F]$ (which of course is equal to the first two terms) to remind you that it is the one form obtained by lowering the index with the metric.  It is easy to verify that the $0$-form $*d*G$ is $\nabla\cdot G$.  To get the fourth line I used that the Lie derivative acts as a derivation on inner products, and the final line used the fact that the exterior derivative of an $n$-form (i.e. $d\epsilon$) must vanish.